The number of molecules of $S O_{2}$ present in $64$ g of $S O_{2}$ is:

6.0 \times 10^{23}

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3 \times 10^{23}

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12 \times 10^{23}

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3 \times 10^{10}

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Solution

The correct option is A $6.0 \times 10^{23}$
The molar mass of $S O_{2}$ is $32 + 16 + 16 = 64 g / m o l$ .
The number of moles of $S O_{2}$ present in $64$ g of $S O_{2}$ is $\frac{64}{64} = 1 m o l$ .
1 mole of any substance contains avogadro's number of molecules which is $6.0 \times 10^{23}$ .
Thus, the number of molecules of $S O_{2}$ present in $64$ g of $S O_{2}$ is $6.0 \times 10^{23}$ .

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The number of molecules of SO2 present in 64 g of SO2 is:

The correct option is A 6.0×1023The molar mass of SO2 is 32+16+16=64g/mol.The number of moles of SO2 present in 64 g of SO2 is 6464=1mol.1 mole of any substance contains avogadro's number of molecules which is 6.0×1023.Thus, the number of molecules of SO2 present in 64 g of SO2 is 6.0×1023.

The number of molecules of $S O_{2}$ present in $64$ g of $S O_{2}$ is: